# Show that if $\sum_{k=1}^{\infty}(u_k)^2 \lt 1$ then $\sum_{k=1}^{\infty}\frac{2^{-k}}{1+3^{-k}-u_k} \lt \infty$

I need to show that if $$\sum_{k=1}^{\infty}(u_k)^2 \lt 1$$ then $$\sum_{k=1}^{\infty}\frac{2^{-k}}{1+3^{-k}-u_k} \lt \infty$$.

I tried using all convergence tests, but to no avail. There's probably some way to decompose the fraction so that you can directly use the convergence of the $$u_i$$ series, but I can't find that.

Can someone maybe help me please?

• The limit comparison test works: Compare with $2^{-k}$ and use $u_k\to 0.$ – zhw. Nov 20 '18 at 17:31

Note that the convergence of $$\sum_{k=1}^{\infty}(u_k)^2$$ implies that $$u_k\to 0$$. Hence $$3^{-k}-u_k\to 0$$ and there is $$N>0$$ such that for all $$k\geq N$$, $$|3^{-k}-u_k|<1/2$$. Then $$\frac{2^{-k}}{1+1/2}\leq \frac{2^{-k}}{1+3^{-k}-u_k}\leq \frac{2^{-k}}{1-1/2}.$$ Can you take it from here?
P.S. Since $$\sum_{k=1}^{\infty}(u_k)^2<1$$ then $$|u_k|<1$$ for each $$k\geq1$$, and it follows that the denominator $$1+3^{-k}-u_k$$ is always positive.